To determine the number of moles of iron (\( \text{Fe} \)) produced from the reaction with 2.50 moles of iron(III) oxide (\( \text{Fe}_2\text{O}_3 \)), let's analyze the stoichiometry of the reaction:
The given reaction is:
[tex]\[ \text{Fe}_2\text{O}_3 + 2\text{Al} \rightarrow \text{Al}_2\text{O}_3 + 2\text{Fe} \][/tex]
From the reaction, we see that:
- 1 mole of \( \text{Fe}_2\text{O}_3 \) produces 2 moles of \( \text{Fe} \).
Given that we start with 2.50 moles of \( \text{Fe}_2\text{O}_3 \), we can use the molar ratio to find the number of moles of \( \text{Fe} \) produced.
1 mole of \( \text{Fe}_2\text{O}_3 \) produces 2 moles of \( \text{Fe} \), so:
[tex]\[
\text{Moles of Fe} = \text{moles of Fe}_2\text{O}_3 \times 2
\][/tex]
Substituting 2.50 moles of \( \text{Fe}_2\text{O}_3 \) into the equation:
[tex]\[
\text{Moles of Fe} = 2.50 \times 2
\][/tex]
[tex]\[
\text{Moles of Fe} = 5.00
\][/tex]
Thus, if 2.50 moles of \( \text{Fe}_2\text{O}_3 \) react, 5.00 moles of iron (\( \text{Fe} \)) will be produced.
[tex]\[
\text{Answer: } 5.00 \text{ moles of Fe}
\][/tex]
